## Step 1: Write the Production Function The number of holes you can dig is given by: \[ Q = (L^{-1} + 9K^{-1})^{-1} \] where: - \( Q \): Number of holes - \( L \): Labor input - \( K \): Capital input ## Step 2: Express Inputs Needed for 10 Holes Set \( Q = 10 \): \[ 10 = (L^{-1} + 9K^{-1})^{-1} \] Take the reciprocal of both sides: \[ 10^{-1} = L^{-1} + 9K^{-1} \] \[ \frac{1}{10} = \frac{1}{L} + \frac{9}{K} \] ## Step 3: Write the Cost Function The cost of using labor and capital is: \[ C = wL + rK \] where: - \( w = \$10 \) (wage rate) - \( r = \$90 \) (price of capital) So, \[ C = 10L + 90K \] ## Step 4: Set Up the Cost Minimization Problem Minimize \( C = 10L + 90K \) subject to \( \frac{1}{L} + \frac{9}{K} = \frac{1}{10} \). Set up the Lagrangian: \[ \mathcal{L} = 10L + 90K + \lambda \left(\frac{1}{L} + \frac{9}{K} - \frac{1}{10}\right) \] ## Step 5: Find the First-Order Conditions Take derivatives and set to zero: ### With respect to \( L \): \[ \frac{\partial \mathcal{L}}{\partial L} = 10 - \lambda L^{-2} = 0 \implies \lambda = 10 L^{2} \] ### With respect to \( K \): \[ \frac{\partial \mathcal{L}}{\partial K} = 90 - 9\lambda K^{-2} = 0 \implies \lambda = 10 K^{2} \] ### Equate the two expressions for \( \lambda \): \[ 10L^2 = 10K^2 \implies L^2 = K^2 \implies L = K \] (Assume positive input, so \( L = K \)) ## Step 6: Substitute \( L = K \) into the Constraint Recall: \[ \frac{1}{L} + \frac{9}{K} = \frac{1}{10} \] With \( L = K \): \[ \frac{1}{L} + \frac{9}{L} = \frac{1}{10} \implies \frac{10}{L} = \frac{1}{10} \implies L = 100 \] So \( K = 100 \). ## Step 7: Find the Minimum Cost \[ C = 10L + 90K = 10(100) + 90(100) = 1,000 + 9,000 = \$10,000 \] --- ## **Final Answers** **How would you dig 10 holes?** - Use **100 units of labor** and **100 units of capital**. **How much would it cost?** - **Total cost = \$10,000**.